I have something that is not very satisfying because it implies that the villagers are very gullible, but it doesn't require magically or hypnotically removing information they already have. Just updating it in a way you control, so that every time they change their mind, they actually have more inf...

Actually that is a completely valid command to give. You can ask the villagers to forget and remember any bits of information on a whim so long as it's in the form of command and follows in accordance with the three rules set for them I have a real problem with this. Forgetting something on command...

There's something that I don't get here. Please consider this first theoretical situation: The goddess actually is just a phantom flying around the village. She is not specifically linked to any of the villagers. The villagers think that she is walking among them, but they're wrong. apart from that,...

Well, I was right - there is a much more elegant maximal strategy. You don't even need to divide it up into cases to use the strategy (though checking it will divide into cases)! Check the players that are first, second, and fourth from your left. Guess whichever color hat is worn by an even number...

Proof of a theoretical maximum, assuming a deterministic strategy: EDIT: Actually, I don't even think we need to assume a deterministic strategy. I'll edit my argument in. There are 128 configurations of hats. Each player, in any given deterministic strategy, will be right in 64 cases and wrong in ...

An absurd strategy where the prime-numbered players (2,3,5,7) pick the minority color and the other players pick the majority color As I said in the OP, "They can devise a strategy, but it can't be nominative. The player's can't take into account their own identity, the whole strategy has to b...

Here's a hat puzzle inspired by this one . Seven perfectly rational players stand in a circle. Everyone gets assigned, randomly and independently, a black hat or a white hat. Nobody can see the color of his own hat but the colors of all other players. All the players must try to guess the color of t...

I don't have a formal proof that this will work for any n>2, but it worked smoothly for the values I tried, and it seems that increasing n just gives more margin of error when you build the strategy. Here it is: First, I have to determine in advance the lower bound of the number of correct guesses. ...

It may be possible to do better with two rules than SirGabriel has with one, but your posted win rate is less than his. To keep the thread alive, I think it is relevant to confirm that it is indeed possible to do better than SirGabriel, under the assumptions pointed out by jaap. For any number of p...

I think I have an O(log(N)) solution. Let A initially be {1, 2, 3, ..., N-1}. At all times, we know that at least one of the numbers is in A. Divide A into three roughly equal sets A 1 , A 2 , and A 3 , and ask about A 1 ∪ A 2 , A 1 ∪ A 3 , and A 2 ∪ A 3 . If we get a "yes" response to on...

Ok, I took a chance to work on my solution again, even if now it is just a correction of Nitrodon's solution, but here is where I landed: Let the initial state be state I, where we have one set containing both numbers. We cut the set in n parts and test every pair of parts. If there is only one yes,...

I think I have an O(log(N)) solution. Let A initially be {1, 2, 3, ..., N-1}. At all times, we know that at least one of the numbers is in A. Divide A into three roughly equal sets A 1 , A 2 , and A 3 , and ask about A 1 ∪ A 2 , A 1 ∪ A 3 , and A 2 ∪ A 3 . If we get a "yes" response to on...

So we just have to show that for any distinct a, b, c and d we can find g, h, i, j and T with a and b but neither c nor d in S(g,h,i,j,T). Let g be the position of any digit where a and c differ, h the position of a digit where a and d differ, i the position of a digit where b and c differ ...

So we just have to show that for any distinct a, b, c and d we can find g, h, i, j and T with a and b but neither c nor d in S(g,h,i,j,T). Let g be the position of any digit where a and c differ, h the position of a digit where a and d differ, i the position of a digit where b and c differ ...

Change your definition of x to this. x takes just one input, and: x(p) internally uses h and works this way: -if program p halts with input p, loop forever -if program p does not halt with input p, return 1 Now x(x) halts if and only if x(x) does not halt. There is an important point here: this is ...

It sounds like you're not familiar with quines , which are programs that can print their own source code. Well, I actually am, wrote one years ago in C, discovered recently that it was possible with compressed files too. The very last part is the one that really blew my mind. I mean, the fact that ...

I'm not sure where you got this proof from, but the explanation on Wikipedia seems pretty clear to me (and different from yours). Well, from various places, including the wikipedia article. Can you tell me how it is different from mine ? Wikipedia's h is the same as my h, my x(i,i) would be wikiped...

For example, you could be passing x as a string (the source code), and x-as-a-program runs some checks on the string to decide whether it halts or not. Ok, but with what inputs? By definition, x halts or not depending on its input. Here, x is supposed to decide wether x halts or not. But with which...

Hello I apologize in advance for how trivial this may seem for anyone that truly understands the halting problem and/or if this question has already been answered here, but my searches were not conclusive. There is something I don't get in the halting problem, and none of the explanations I've read ...

The two pairs of integers have the same sum, so they have the same arithmetic mean, x.The first two numbers are x-10 and x+10, the two other are x-13 and x+13.The first product is x²-100, the second one is x²-169.The difference between the products is 69.

Of course I solved tis with the classic method, but considering how some of the statements were designed, I wanted to try something else. I consider the nature of a statement depends only on how this statement matches reality, and not on what other statements say about it. For example "This sta...

Does Alice have a way of telling Bob the transfer is over, or do we need to establish a protocl which allows Bob to know when he has read the last bit, and stop expecting new information from Alice ? If Bob can know when the transfer is over wothout having to figuring it out with the data, then we c...

you need to tell the weights beforehand. Just checking for clarification of this point. Does it mean that you know what weights both types of coins are before you start weighing them (eg: you know the real coins are (say) 30g and the fake ones are (say) 29 grams)? Or does it mean something else? I'...

I have never seen this variation of the classic balance puzzles before, so I tried to solve it and found the solution. I am not going to tell a story around this because it is a classic type of problem, the deal is that you have 7 coins, but 2 of these coins are slighly lighter than the other 5. Of ...

Even given superrationality, I disagree with the superrational solutions posted so far... Say every one of the million people chooses to play with probability p. Then we would, between them all, expect a total of 10 8 p dollars be paid in entry fees, and a total of 10 6 (1 - (1-p) 10000...

I totally forgot about mixed strategies: if the perfectly logical strategy is to play with a probability p, then there are 1,000,000p players to split the $1,000,000, so the gain is 1/p-100. As this happens with probability p, that makes p(1/p-100), that is 1-100p. That means that if p is be...

If all other players are perfect logicians, then they will all come to the same conclusion : "Either I play and everybody plays, and we all lose $99, or I don't play, and nobody plays, and we are all even. I don't play." Now, I am not a perfect logician, I am just human, so my reas...

Two solutions which came to my mind: 1) Use an asymmetric encryption method. Each member i has a public key A_i which is known by all, and a private key B_i only known to them. In a meeting of 2 persons, both exchange their number i. Both generate a random string, encypt it with the public key ...

tomtom2357 wrote:Okay. what is the highest percentage of the rectangle you can use, I have 2/3 so far.

I get 3/4 : draw crosses (connecting opposite corners) on two opposite faces, cut along the lines, then cut one of the edges linking those two faces. Other cuttings of the opposite faces get the same result.

My attempt was based on what felt like the most obvious solution, assuming the discs were not required to be each on one side of the rectangle. It consists of one vertical rectangle 29.7cm long, this detemines the radius of the discs, both discs are put on the same side of the rectangle, thus determ...

Imagine you are one of the prisonners, and you see 27 red hats, 31 white hats and 41 blue hats. The value you see is 31+2*41 = 113 (congruent to 2 modulo 3). Now you are lucky and you are not the first one to be picked. The guy has a white hat, so the value he sees is 112 + the value of your own hat...